Distance between two points 

Before we discuss the procedure of finding the distance between points, we shall first throw some light on the position of two points with respect to a given line. These concepts lay the foundation for various other concepts and hence are extremely useful for attempting numerical.

Consider a point (x1, y1) and a line ax + by + c = 0. Then, if ax1 + by1+ c is of the same sign as ‘c’, then the point (x1, y1) lie on the same origin side of ax + by + c = 0 . Conversely, if the sign of c and ax1 + by1+ c is opposite, then the point (x1, y1) will lie on the non-origin side of ax + by + c = 0.

Let the line be ax + by + c = 0 and P(x1, y1), Q(x2, y2) be two points. Then, we have two cases:

Case 1:

If P(x1, y1) and Q(x2, y2) are on the opposite sides of the line ax + by + c = 0, then the point R on the line ax + by + c = 0 divides the line PQ internally in the ratio m1 : m2, where m1/m2 must be positive.

Co-ordinates of R are {(m1x+ m2x1)/(m+ m2), (m1y+ m2y1)/(m+ m2)}.

Point R lies on the line ax + by + c = 0.

⇒ m1/m2 = (ax+ by+ c)/(ax+ by+ c) > 0

1458_Position of two points with respect to a given line.JPG

So that ax1 + by1 + c and ax2 + by2 + c should have opposite signs.

Case 2:

If ax1 + by1 + c and ax2 + by2 + c have the same signs then m1/m2 = –ve, so that the point R on the line ax + by + c = 0 will divide the line PQ externally in the ratio m1 : m2 and the points P(x1, y1) and Q(x2, y2) are on the same side of the line ax + by + c = 0.

Illustration:

Find the range of θ in the interval (0, π) such that the points (3, 5) and (sin θ, cos θ) lie on the same side of the line x + y – 1 = 0.

Solution:

As discussed above, we know that ax1 + by1 + c and ax2 + by2 + c both must be of the same signs for the points to be on the same side of the line.

Now, 3 + 5 – 1 = 7 > 0 ⇒ sin θ + cos θ – 1 > 0

⇒ sin(π/4 + θ) > 1/√2

⇒ π/4 < π/4 + θ < 3π/4

⇒ 0 < θ < π/2.

Hence, this is the required range of θ in the interval (0, π) such that the points (3, 5) and (sin θ, cos θ) lie on the same side of the line x + y – 1 = 0.

___________________________________________________________________________

Illustration:

Find α, if (α, α2) lies inside the triangle having sides along the lines 2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.

Solution:

Let A, B, C be vertices of the triangle.

A ≡ (–7, 5), B ≡ (5/4, 7/8) and C ≡ (1/3, 1/9).

Sign of A w.r.t. BC is –ve.

1566_triangle having sides along the lines.JPG

If P lies inside the ∆ABC, then sign of P will be the same as sign of A w.r.t. the line BC

⇒ 5α – 6α2 – 1 < 0.                     …… (1)

Similarly 2α + 3α2 – 1 > 0.          …… (2)

And, α + 2α2 – 3 < 0.                  …… (3)

Solving, (1), (2) and (3) for α and then taking intersection,

We get α ∈ (1/2, 1) ∪ (–3/2, –1).

_________________________________________________________________________________________________

Illustration:

The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are respectively x – y + 5 = 0 and x + 2y = 0. If the co-ordinates of A are (1, –2), find the equation of BC.

Solution:

From the figure, since OE and OF are the perpendicular bisectors of the sides AB and AC, so we have the coordinates of points E and F as

E ≡ (x+ 1/2, y- 2/2),

F ≡ (x+ 1/2, y- 2/2).

59_equations of the perpendicular bisectors of sides of triangle.JPG

Since E and F lie on OE and OF respectively,

x1 – y1 + 13 = 0                                  … (1)

and x2 + 2y2 – 3 = 0                           … (2)

Also, slope of AB = –1 and slope of AC is 2, so that

x1 + y1 + 1 = 0.                                   … (3)

And 2x2 – y2 – 4 = 0                           … (4)

Solving these equations, we get the co-ordinates of B and C as

B ≡ (–7, 6) and C ≡ (11/5, 2/5)

⇒ Equation of BC is 14x + 23y – 40 = 0.

____________________________________________________________________________

Illustration:

Two fixed points A and B are taken on the co-ordinate axes such that OA = a and OB = b. Two variable points A’ and B’ are taken on the same axes such that OA’ + OB’ = OA + OB. Find the locus of the point of intersection of AB’ and A’B.

Solution:

Let A ≡ (a, 0), B (0, b), A’ ≡ (a’, 0), B’ ≡ (0, b’).

Equation of A’B is x/a' + y/b' = 1.                                          …. (1)

and equation of AB’ is x/a + y/b'  = 1.                                   …. (2)

Subtracting (1) from (2), we get, x (1/a – 1/a') + y(1/b' – 1/b) = 0.

⇒ x(a'–a)/aa' + y(b–b')/bb' = 0.                   [Using a’ – a = b – b’]

⇒ x/a(b–b'+a) + y/bb' = 0 

⇒ b’ = a(a+b)y/ay–bx.                   ….. (3)

From (2), b’x + ay = ab’   ….. (4) we get x + y = a + b which is the required locus.

The distance between two points P(x1, y1) and Q(x2, y2) is (see the figure given below).

198_Distance between two points 1.JPG

Length PQ = √(x2 – x1)2 + (y2 – y1)2

Proof:

Let P(x1, y1) and Q(x2, y2) be the two points and let the distance between them be ‘d’. Draw PA, QR parallel to y-axis and PR parallel to x-axis. 

Angle QRP = 90o  

⇒ d2 = PR2 + RQ2 

⇒ d2 = (x2 – x1)2 + (y2 – y1)2

⇒ d = √(x2 – x1)2 + (y2 – y1)2

You may also view the video on distance formula

Similarly, in three dimensions, we shall have a third coordinate and the formula in that case changes a bit:

Distance between P(x1, y1, z1) and Q(x2, y2, z2) is given by the formula d = √(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2.

Let us say we want to know the co-ordinates of point which divides a line segment between two points A(x1, y1) and B(x2, y2) in the ratio m : n. Then two types of division are possible:

The coordinates of such a point are given by:

Internal Division: {(nx1 + mx2)/(m + n), (ny1 + my2)/(m + n)}

External Division:  {(mx2 - nx1)/(m - n), (my2 - ny1)/(m - n)}

Note: 

This is called section formula. 

Let P divide the line segment AB in the ratio m : n. If P is inside AB then it is called internal division; if it is outside AB then it is called external division. 

However in each case AP/BP [or AP'/BP' or AP"/BP"] = m/n.  

Proof: 

Consider ∆ ABB’  

Since BB’||PQ and AP:PB = m:n (see figure given below)

2370_Distance between two points 2.JPG

AQ/AB' = PQ/BB' = m/(m+n) (= AP/AB) 

⇒ x – x1/x2 – x1 = m/(m+n) 

⇒ x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n) 

If P is outside AB (less assume it is at P’) 

We have (x – x1)/(x2 – x1) = m/(m + n) 

⇒ x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n) 

Similarly if P is at P” then 

x = (–mx+ nx1)/(n-m), y = (– my+ ny1)/(n-m)

Note:  

m:n can be written as m/n or λ:1. So any point on line joining A and B will be P{(λx+ x1)/(λ + 1), (λy+ y1)/(λ +1)}. It is useful to assume λ:1 because it involves only one variable. 

____________________________________________________________________________

Illustration: 

Prove that altitudes of a triangle are concurrent and prove that the co-ordinates of the point of con-currency are {(x1 tan A + x2 tan B + x3 tan C)/(tan A + tan B + tan C), (y1 tan A + y2 tan B + ytan C)/(tan A + tan B + tan C)}. 

Solution: 

In triangle A(x1, y1), B(x2, y2) and C(x3, y3), draw AD perpendicular to BC. Our effort now should be to find the co-ordinates of the point D.

178_Distance between two points 3.JPG

To do that, we need to find BC/CD. (figure is given above) 

tan B = AD/BD and tan C = AD/CD 

⇒ BD/DC = tan C/tan B 

Now we apply section formulae: 

xD = (x2 tan B + x3 tan C)/(tan B + tan C)                           …… (i)

yD = (y2 tan B + y3 tan C)/(tan B + tan C)                               …… (ii)

We know that orthocenter will lie on AD. We need to find this point and its co-ordinates. 

We should select a point H1 on AD and take the ratio AH1/H1D in such a manner so that xH1 and yH1 calculated form (i) should be symmetric in x1, x2, x3, tan A, tan B and tan C.

Let AH1/H1D = (tan B + tan C)/tan A 

⇒ xH1 = (x1 tan A + x2 tan B + x3 tan C)/(tan A + tan B + tan C)

and ⇒ yH1 = (y1 tan A + y2 tan B + y3 tan C)/(tan A + tan B + tan C) 

Since the result is symmetric, this point H1 will lie on other altitude as well i.e. the altitudes are concurrent

⇒ xH = xH1 and yH = yH1.

___________________________________________________________________________

Illustration: 

Prove analytically that in a right angled triangle the midpoint of the hypotenuse is equidistant from the three angular points. 

Solution: 

While proving a problem analytically take most convenient co-ordinates of known points. 

In the present case triangle is assumed as AOB with coordinates as shown in figure given below, C is midpoint of AB.

236_Distance between two points 4.JPG

 

 

 

 

 

 

 

So co-ordinates of C will be (a/2, b/2) 

Now AB = √a2 + b2

CA = CB = AB/2 (C is mid point of AB) 

= √a2 + b2 

And, we know that the distance between two points C and O is given by 

CO = √(a/2 – 0)2 + (b/2 – 0)2 = √a2 + b2/2 

Hence CA = CB = CO 

Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) internally in the given ratio λ1 : λ2 i.e., AP/BP = λ12 are P {(λ2x+ λ1x2)/(λ+ λ1), (λ2y+ λ1y2)/(λ+ λ1)}. 

Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) externally in the ratio λ1 : λ2 i.e., AO/BP = λ1/λ2 are P(λ2x1+λ1x2)/(λ2λ1), (λ2y1+λ1y2)/(λ2λ1).

Related Resources:

To know more about the study material of engineering and medical exams, please fill up the form given below:

Name
Email Id
Mobile

Exam
Target Year

Related Resources
Examples on Distance between two Parallel Lines

Examples on Distance between two Parallel Lines...

Centroid Incentre and Circum Centre

Centroid, Incentre, Circumcentre and Ex-centre of...

Standard Equations of Straight Line

Straight Line Any equation of first degree of the...

Concurrency of Straight Lines

Concurrency of Straight Lines What do you mean by...

Angle Bisectors

Angle Bisectors The concept of angle bisector is...

Family of Lines

Family of Lines Straight lines is an important...

Distance between two Parallel Lines

Distance between two Parallel Lines In this...

Combined Equations of Angle Bisectors of Lines

Combined Equations of Angle Bisectors of Lines...

Angle between two Straight Lines

Angle between two straight lines Angle bisector of...

Angle between Pair of Lines

Angle between Pair of Lines Straight lines is an...

Position of two Points with Respect to a given Line

Position of two points with respect to a given...

Different Forms of Line

Different Forms of Line Straight line is an...

Joint Equation of Pair of Lines

Joint Equation of Pair of Lines Joint Equation of...

Area of Triangle

Area of a triangle Let (x 1 , y 1 ), (x 2 , y 2 )...

Examples based on Straight Line

Examples based on Straight Line Straight lines is...

Representation of Points in Plane

Representation of Points in a Plane Coordinate...

Pair of Straight Lines

Pair of Straight Lines The equation ax 2 + 2hxy +...

Examples on Angle between two Straight Lines

Examples on Angle Between two straight lines....